Question: What is the value of $\dfrac{d}{dx}\left(\dfrac{1}{x^5}\right)$ at $x=2$ ?
Explanation: The strategy We can first rewrite the fraction as a negative power of $x$. Then, the derivative can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=2$. Rewriting the fraction as a negative power $\dfrac{1}{x^5}=x^{-5}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(x^{-5}\right) \\\\ &=-5x^{-5-1} \gray{\text{The power rule}} \\\\ &=-5x^{-6} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\dfrac{1}{x^5}\right)=-5x^{-6}$, which can also be written as $-\dfrac{5}{x^6}$. Now let's plug ${x=2}$ : $\begin{aligned} -\dfrac{5}{({2})^6}&=-\dfrac{5}{64} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{1}{x^5}\right)$ at $x=2$ is $-\dfrac{5}{64}$.